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danrussekme

Answered question

2021-12-04

A man throws a ball into the air with a velocity of $96 \frac{f t}{s}$ Use the formula $h = - 16 t^{2} + v_{0} t$ to determine when the height of the ball will be 48 feet. Round to the nearest tenth.

Answer & Explanation

giskacu

Beginner2021-12-05Added 22 answers

Step 1
Given:
Velocity $v_{0} = 96 \frac{f t}{s}$
Use formula $(h) = - 16 t^{2} + v_{0} t$
Step 2
$\Rightarrow$ To determine when the height of ball will be 48 feet
$\Rightarrow h = - 16 t^{2} + v_{0} t$
Substitute $v_{0} = 96 \frac{f t}{s}$
$∴ h = - 16 t^{2} + 96 t$
$∴ 48 = - 16 t^{2} + 96 t$
$\Rightarrow h = 48 f e e t$
$\Rightarrow 16 t^{2} - 96 t + 48 = 0$
$16 (t^{2} - 6 t + 3) = 0$
$t^{2} - 6 t + 3 = 0$
We know that:
Quadratic formula for $a x^{2} + b x + c = 0$
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$
$\Rightarrow t = \frac{- (- 6) \pm \sqrt{{(- 6)}^{2} - 4 (1) (3)}}{2 (1)}$
$= \frac{6 \pm \sqrt{36 - 12}}{2}$
$= \frac{6 \pm \sqrt{24}}{2}$
$= \frac{6 \pm 2 \sqrt{6}}{2}$
$= \frac{2 (3 \pm \sqrt{6})}{2}$
$t = 3 \pm \sqrt{6}$
$\Rightarrow t = 3 + \sqrt{6}; t = 3 - \sqrt{6}$
$t = 5.449$ ;
$t = 0.5505$
$∴ t = 5.4 \sec$
$t = 0.6 \sec$
$∴$ The height of ball will be 48 feet at t 54 sec and $t = 0.6 \sec$

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